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Suppose I have the following RF mixer circuit-Single diode RF mixer

How does placing two AC voltages of differing magnitude and frequency directly in contact with one another not create a short circuit condition in a circuit like this? By short circuit condition, I mean placing two different voltage potentials directly in contact with one another. I realize this circuit works in reality when built and tested, but this question has always bothered me from a theoretical perspective, even beyond encountering circuits like this (e.g., DC biasing transistors with AC inputs as well). I know that at least in the case of DC biasing transistors, superposition seems to be the correct answer. But it still seems intuitively like you are directly combining two different potentials, and therefore creating a short circuit condition.

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    \$\begingroup\$ Usually the source feeding an effort device is not an ideal voltage source but a 50 ohm (or some other chosen impedance) source. \$\endgroup\$
    – The Photon
    Jan 15 '18 at 16:01
  • \$\begingroup\$ I think that the underlying assumption is that the RF and LO are ac coupled. \$\endgroup\$
    – Mike
    Jan 15 '18 at 16:19
  • \$\begingroup\$ @The Photon, does this mean that you can view it as two AC signals with series resistances adding together via superposition? \$\endgroup\$
    – user468756
    Jan 15 '18 at 18:57
  • \$\begingroup\$ At least it means the LO input doesn't short out the RF input. The net effect will be some kind of superposition of the two inputs appearing on the diode's anode. \$\endgroup\$
    – The Photon
    Jan 15 '18 at 19:12
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If the sources (RF input and LO input) are current sources connecting them together in a node means adding the currents.

Since the diode afterwards is a device with a highly non-linear \$v\$-\$i\$-relationship the voltage afterwards will not be a linear function of the sum of the source currents \$i_1 + i_2\$ but it will contain higher order terms:

The \$v\$-\$i\$-relationship can be approximated in the vicinity of the operating point by a power series:

\$v = c_0 + c_1 (i_1 + i_2) + c_2 (i_1 + i_2)^2 + c_3 (i_1 + i_2)^3 + ...\$

(for a linear device (e.g. resistor, capacitor, inductor) only the linear coefficient \$c_1\$ is non-zero, all others are 0;
for a non-linear device (like a diode), also coefficients of non-linear terms are non-zero)

If you multiply out the non-linear terms you get the mixing products, e.g. \$i_1^2\$, \$i_1 i_2\$, \$i_2^2\$, \$i_1^3\$, \$i_1^2 i_2\$, \$i_1 i_2^2\$ , \$i_2^3\$, ....

The unwanted mixing products (e.g. anything but \$i_1i_2\$) are removed by the LC tank circuit at the IF output.

EDIT: Explaining why it works also with two non-ideal voltage sources:

schematic

simulate this circuit – Schematic created using CircuitLab

The left diagram has two Norton (current) sources, the right diagram has two Thevenin (voltage) sources and both circuits are equivalent, i.e. they cause the same current in the load (even if the load was not a resistor but a non-linear element).

And looking at the left circuit the current would be:
\$i_L = (i_1 + i_2) \frac{R_L + R_{12}}{R_L}\$ (current divider) where \$R_{12}=\frac{R_1 R_2}{R_1 + R_2}\$

The important result is that also in this circuit using non-ideal current sources (or equivalently using non-ideal voltage sources) the resulting current is some constant times the sum of the currents of the sources (or sum of the voltages of the sources).

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  • \$\begingroup\$ The sources don't have to be current sources, though, right? \$\endgroup\$
    – user468756
    Jan 15 '18 at 18:58
  • \$\begingroup\$ It easiest to explain and understand if you assume ideal current sources. Of course it works also with Norton sources (i.e. an ideal current source shorted by a resistor) and for those there exist equivalent Thevenin sources (ideal voltage sources with a resistor in series). So you can use two voltage sources with a series resistor each (e.g. 50 Ohm sources). It works not, however, with ideal voltage sources. As you said correctly: connecting them together would cause a short circuit. \$\endgroup\$
    – Curd
    Jan 15 '18 at 19:36
  • \$\begingroup\$ @user468756: see also my EDIT answering your question in more detail \$\endgroup\$
    – Curd
    Jan 15 '18 at 21:51
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Your circuit has substantial PI-network filtering in the RF path.

If the LO path has similar filtering, then some form of linear superposition occurs.

In truth, the total capacitance on the diode's anode must be including in computing the final inductors in each of RF and LO PI chains, to maintain the resonances and energy transfers.

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